Question

The roots of the equation $4^x-3\cdot 2^{x+3}+128=0$ are

• A. 4 and 5
• B. 3 and 4
• C. 2 and 3
• D. 1 and 2

Answer 1

Answer: (B)

We have, $4^x-32^{x+3}+128=0$
$\Rightarrow 2^{2x}-3\cdot 2^x\cdot 2^3+128=0$
$\Rightarrow 2^{2x}-24\cdot 2^x+128=0$
$\Rightarrow y^2-24y+128=0$ where $2^x=y$
$\Rightarrow (y-16)(y-8)=0$
$\Rightarrow y=16,8$
$\Rightarrow 2^x=16$ or $2^x=8$
$\Rightarrow x=4$ or 3